Find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6).

1 answer:

3 0

hello :
the parabola's equation is : f(x) = a(x-1)²-9
the coefficient of the squared expression in the parabola's equation is : a
but ; f(0) = -6 ... ( y-intercepts (0,-6) )

-6 = a(0-1)²-9
-6 =a -9

a=3
f(x) = 3(x-1)²-9

the x-intercepts when : f(x) =0
3(x-1)²-9 =0
(x-1)² =3
x-1 = √3) or x-1 = -√3)

x= 1 +√3 or x= 1 -√3 are the x-intercepts

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The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

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$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$